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Horrible mechanics :(

Hi, I've got this horrible problem and can't get near the 'show' solution. I apologise for my notes being quite messy but here's what I've done essentially:

1) Consider the vertical direction and find an equation for V. (2)
2) Consider horizontal direction and use speed=distance/time to get D=Vcosa*(t1+t2)
3) Consider vertical direction to find t1 (time to go up) and t2 (time to go down) and put this into (1)
4) Sub V from (2) into this new equation.

I must be missing something somewhere?

Thanks for any pointers!
sadmech.png19.6KB
IMG_3398.jpg474.2KB
Original post by Substitution
Hi, I've got this horrible problem and can't get near the 'show' solution. I apologise for my notes being quite messy but here's what I've done essentially:

1) Consider the vertical direction and find an equation for V. (2)
2) Consider horizontal direction and use speed=distance/time to get D=Vcosa*(t1+t2)
3) Consider vertical direction to find t1 (time to go up) and t2 (time to go down) and put this into (1)
4) Sub V from (2) into this new equation.

I must be missing something somewhere?

Thanks for any pointers!


Perhaps @TeeEm or @Zacken can help

But I got H+D2tan⁡2αH+Dtan⁡αH + \dfrac{D^2 \tan ^2 \alpha}{H+ D \tan \alpha} when I did the question. I'll look through my working but not sure where the random 4 is coming from.

Edit: in my working I said that 2x9.8=4.9 for some reason. I'll try to help.

What I did:

•

Look at it horizontally to find out t.

•

Look at it vertically to find out V^2 using t from the first step.

•

Look at it vertically to work out the height between the apex and starting point. I call this height m. Use v^2 value from the previous step.

In the first part of your working you have said that u=Vsin⁡αu=V \sin \alpha but then have said u2=V sinαu^2=V\ sin \alpha. This does not follow. Perhaps those I've quoted in will spot other errors but I'm afraid I'm not following your working :colondollar:.

I'll spoiler in what I've done but I'm sure there are other solutions. Hopefully you can solve it using the help you'll get from the others :h:.

Spoiler

(edited 8 years ago)
Original post by Kvothe the arcane
Perhaps @TeeEm or @Zacken can help

But I got H+D2tan⁡2αH+Dtan⁡αH + \dfrac{D^2 \tan ^2 \alpha}{H+ D \tan \alpha} when I did the question. I'll look through my working but not sure where the random 4 is coming from.

Edit: in my working I said that 2x9.8=4.9 for some reason. I'll try to help.

What I did:

•

Look at it horizontally to find out t.

•

Look at it vertically to find out V^2 using t from the first step.

•

Look at it vertically to work out the height between the apex and starting point. I call this height m. Use v^2 value from the previous step.

In the first part of your working you have said that u=Vsin⁡αu=V \sin \alpha but then have said u2=V sinαu^2=V\ sin \alpha. This does not follow. Perhaps those I've quoted in will spot other errors but I'm afraid I'm not following your working :colondollar:.

I'll spoiler in what I've done but I'm sure there are other solutions. Hopefully you can solve it using the help you'll get from the others :h:.

Spoiler



I am so glad to get the answer to this (looked at spoiler). It's been driving me crackers.
Original post by maggiehodgson
I am so glad to get the answer to this (looked at spoiler). It's been driving me crackers.


I'm glad my solution helped you :smile:.

Did you follow? Could you start from a different place and get the result now?
Original post by Kvothe the arcane
I'm glad my solution helped you :smile:.

Did you follow? Could you start from a different place and get the result now?


HI

Yes, I did follow it and No I couldn't start from a different place. Wouldn't know which place!!!
Reply 5
Avatar for TeeEm
TeeEm
19
Original post by Substitution
Hi, I've got this horrible problem and can't get near the 'show' solution. I apologise for my notes being quite messy but here's what I've done essentially:

1) Consider the vertical direction and find an equation for V. (2)
2) Consider horizontal direction and use speed=distance/time to get D=Vcosa*(t1+t2)
3) Consider vertical direction to find t1 (time to go up) and t2 (time to go down) and put this into (1)
4) Sub V from (2) into this new equation.

I must be missing something somewhere?

Thanks for any pointers!


If you are still stuck let me know and i will post my version
Original post by TeeEm
If you are still stuck let me know and i will post my version


Hi there. I'm very interested in seeing your solution. Would you mind attaching it - perhaps in a spoiler?
Reply 7
Avatar for TeeEm
TeeEm
19
Original post by Kvothe the arcane
Hi there. I'm very interested in seeing your solution. Would you mind attaching it - perhaps in a spoiler?


here they are

Spoiler

IMG_0002.jpg363.7KB

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